Question

Inordertoobtainamagnificationof-1.5withaconcavemirroroffocal

length16cm,theobjectwillhavetobeplacedatadistanceof

a)between6cm and16cm b)between32cm and16cm c)between48cm

and32cm d)beyond64cm​

Answer 1

Answer:

b)

Explanation:

Answer 2

Answer:

b) between 32cm & 16 cm

Explanation:

-1.5 cm shows that the image is 1.5, times magnified and inverted.

Concave mirror forms enlarged image only in two cases, first between C&f , and second between f&P.

In the first case the image is inverted, and in the second case it's erect.

As given in the question we need inverted image.

so the object should be placed between focus and centre of curvature.

Focal length= 16cm

Radius of Curvature= 2(16) = 32cm.

So object should be kept between 32cm & 16 cm.

Mathematically,

$m = \frac{f}{f - u} \\ - 1.5 = \frac{16}{16 - (-u)} \\ - 24 - 1.5u = 16 \\ -1.5u = 40 \\ u = 40 \div -1.5 \\ = -26.67cm$

We have to keep the object at 26.67cm.(Negative sign shows that it's a position of real object - Using sign convention)

Which is between 32cm & 16 cm